离散数学——Algorithm and Recurrence

目录

• Algorithm
  • Big-O
  • N, NP, NPC, NP Hard
  • Input size
• Recurrence
  • First-order linear recurrence
  • Master Theorem
  • Linear homogeneous relation
  • Linear non-homogeneous recurrence relation

Algorithm

Big-O

Definition of Big-O

These are some frequently used Big-O estimates for some functions

There are rules for combination of Big-O

Definition of function type

Under this definition, polynomials are of the same type, but a polynomial and a exponential aren't

N, NP, NPC, NP Hard

P(polynomial)问题：能在多项式时间内解决的问题

NP(non-deterministic polynomial)问题：能在多项式时间内验证答案的问题

P属于NP，能在多项式时间内解决，一定能在多项式时间验证

NP complete: NP complete 属于 NP，即可以在多项式时间内验证，且所有NP问题都可以归约到NP complete，NP complete是NP中最难的问题，能解决NP complete的算法就能解决所有NP

NP hard: NP hard不一定能在多项式时间内验证，NP hard与NP有交集，交集就是NP complete，所有NP都能归约到NP hard

Input size

Recurrence

First-order linear recurrence

First-order linear recurrence: 形如\(T(n) = f(n)T(n-1) + g(n)\)

解决First-order linear recurrence function，一步步往回推找到规律即可，最后用base case来替代掉T(n)

Master Theorem

Linear homogeneous relation

Linear homogeneous relation is of the form: \(a_n = c_1a_{n-1} + c_2a_{n-2} + ... + c_ka_{n-k}\)

The solution to \(a_n\) is of the form: \(r^n\)

Put the solutions of \(a_n, a_{n-1}, ...\) back to the linear homogeneous relation, and we get characteristic function

For characteristic function of degree 2, we have the following theorem

1. When the function has 2 different roots

2. When the function only has 1 root

Then put the condition we get on base case, then we can solve \(\alpha_1\) and \(\alpha_2\)

For characteristic function of more than 2 degree, we have the following theorem

Linear non-homogeneous recurrence relation

Linear non-homogeneous recurrence relation is of the form: \(a_n = c_1a_{n-1} + c_2a_{n-2} + ... + c_ka_{n-k} + f(n)\)

The solution to \(a_n\) is of the form \(a_n = r^n + p(n)\)

we can get \(r^n\) by ignoring the f(n) part, and use method in solving linear homogeneous recurrence relation

for \(p(n)\), we assume \(p(n) = cn + d\), and now we ignore the \(r^n\) part, and let \(a_n = p(n)\) to solve \(p(n)\)